The probability distribution of the number of hours cars are parked has been estimated as follows.

x 1 2 3 4 5 6 7 8

p(x) .24 .18 .13 .10 .07 .04 .04 .20

The mean is equal to:

1*0.24 + 2*0.18 + 3*0.13 + 4*0.1 + 5*0.07 + 6*0.04 + 7*0.04 + 8*0.2 = 3.86

The standard deviation is:

sqrt((1^2*0.24 + 2^2*0.18 + 3^2*0.13 + 4^2*0.1 + 5^2*0.07 + 6^2*0.04 + 7^2*0.04 + 8^2*0.2) - 3.86^2)

=> sqrt((1*0.24 + 4*0.18 + 9*0.13 + 16*0.1 + 25*0.07 + 36*0.04 + 49*0.04 + 64*0.2) - 3.86^2)

=> sqrt(21.68 - 14.8996)

=> sqrt(6.7804)

=> 2.60392

**The mean of the probability distribution is 3.86 and the standard deviation is approximately 2.60392**

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